# Thread: Working with sec...any thoughts

1. ## Working with sec...any thoughts

Hi,

I'm solving the integral of sec x.

I'm using a t sub ie $t = tan(x/2)$

I've integrated to $ln(|1+t|/|1-t|)$

Now I believe I have to multiply inside the brackets by $cos(x/2)+sin(x/2)$ so I'll end up with

$(cos^2(x/2)+sin^2(x/2)+2cos(x/2)sin(x/2))/cos(x)$

Could someone explain that to me?

Thank You.

2. Originally Posted by minusb
Hi,

I'm solving the integral of sec x.

I'm using a t sub ie $t = tan(x/2)$

I've integrated to $ln(|1+t|/|1-t|)$

Now I believe I have to multiply inside the brackets by $cos(x/2)+sin(x/2)$ so I'll end up with

$(cos^2(x/2)+sin^2(x/2)+2cos(x/2)sin(x/2))/cos(x)$

Could someone explain that to me?

Thank You.
try the substitution

$u=\sec(x)+\tan(x) \implies du = \sec(x)\tan(x)+\sec^2(x)\, dx = \sec(x) u \,dx$

This gives the integral

$\displaystyle \frac{1}{u}\,du=...$

3. Thanks, I know that version... The question is asking to use t=tan(x/2) substitution.

4. Originally Posted by minusb
I've integrated to $ln(|1+t|/|1-t|)$
Well, you have done it all! Just resubstitute $t$ and note that $1 = \tan\frac{\pi}{4}$, then use the tangent addition identity:

$\displaystyle \ln\left\{\frac{1+t}{1-t}\right\}+k = \ln\left\{\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right\}+k = \ln\left\{\frac{\tan{\frac{\pi}{4}}+\tan\frac{x}{2 }}{\tan{\frac{\pi}{4}}-\tan\frac{x}{2}}\right\}+k = \displaystyle \ln\left\{\tan\left(\frac{\pi}{4}+\frac{x}{2}\righ t)\right\}+k.$

5. Originally Posted by TheCoffeeMachine

$\displaystyle \ln\left\{\tan\left(\frac{\pi}{4}+\frac{x}{2}\righ t)\right\}+k.$
Is this a common identity that is equal to secx + tanx?

6. Double post...

7. Originally Posted by minusb
Is this a common identity that is equal to secx + tanx?
Why do you have to show that they are equal? This is the classical integral of $\displaystyle \sec{x}$!
If they have asked you to use that substitution, it should really be what they wanted!